33 Idempotents and Characters

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Janel Black
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1 33 Idempotents and Characters On this day I was supposed to talk about characters but I spent most of the hour talking about idempotents so I changed the title. An idempotent is defined to be an element e of any ring so that e 2 = e. If the ring is an endomorphism ring then e is called a projection and denoted p. I used the following fact from linear algebra without proof. Theorem An n n matrix P with coefficients in a field F satisfies the equation P 2 = P if and only if it is a projection map (not necessarily orthogonal) onto the subspace im P F n along ker P. Furthermore the dimension of im P (the rank of P ) is equal to its trace: dim F im P = Tr P. Proof. Actually, I did give a proof. ( I said ) it was because P is similar (i.e. I conjugate) to a matrix of the form. This follows from the fact that F n = im P ker P. Thus if we choose a basis for im P and a basis for ker P we get a basis for F n and the matrix of P with respect to this basis is as above. Now the theorem follows from the fact that rank and trace do not change when we conjugate a matrix. I explained everything in two examples Example of a Semisimple Algebra By the Wedderburn Theorems a semisimple (s-s) algebra over C is a product of matrix algebras: A = M f1 (C) M fr (C). Let s take an example: A 123 = M 1 (C) M 2 (C) M 3 (C). This is the subalgebra of the matrix algebra given as follows. 1

2 Each star ( ) represents a variable. Since there are 14 stars, this algebra is 14 dimensional. The stars are arranged in squares called blocks. The number of stars is obviously equal to the sum of the squares of the sizes of the individual matrix algebras. 14 = In general dim A = f 2 i. For a group ring A = CG, the dimension is equal to the size of G. This gives: Theorem If CG = M f1 (C) M fr (C) then = r fi 2. i=1 Next, I talked about the significance of the number r (the number of blocks). For the case of group rings it is the number of conjugacy classes in G. This has to do with how many central elements the group algebra has. Going back to our 14 dimensional example, what are the central elements of A 123? Since the central elements of a matrix algebra are the scalar matrices (diagonal matrices with the same diagonal entry): λ λ λ the central elements of A 123 are: λ 1 λ 2 λ 2 λ 3 λ 3 λ 3 There are three eigenvalues, one in each block. Consequently we see that: r = dim Z(M f1 (C) M fr (C)) where Z(A) denotes the center of the algebra A. Theorem An element a g g CG is central iff the coefficients a g are constant on the conjugacy classes of G (i.e., a g = a hgh 1). Consequently, r = dim Z(CG) = # conjugacy classes in G. 2

3 Proof. It is clear that any central element element a g g CG satisfies this condition since ( ag g = h ag g) = a g hgh 1 implies that the coefficients of hgh 1 on both sides agree. But the coefficient of hgh 1 on the left side is a hgh 1 and on the right side it is a g. Thus a hgh 1 = a g. Conversely, if the coefficients of a g g satisfy this condition then they commute with every h G and consequently they commute with all elements of CG. I Among the central elements of an algebra which ones are idempotent? i.e., satisfy e 2 = e? There are only two complex numbers with this property, namely, and 1. Therefore there are 2 3 = 8 central idempotents in our algebra A 123. (In general there are 2 r central idempotents.) They are: 1 e 1 =, e 2 =, e 3 =, 2 e 1 + e 2, e 1 + e 3, e 2 + e 3, 1 = e 1 + e 2 + e 3 and. The first three central idempotents are called primitive because they cannot be written as a sum of two other nontrivial central idempotents. Note that the primitive central idempotents are orthogonal in the sense that e i e j = if i j. [Primitive central idempotents are always orthogonal since otherwise e i = e i (e j + 1 e j ) = e i e j + e i (1 e j ) is a decomposition of e i into a sum of two nontrivial central idempotents.] Note also that 1 = e 1 + e 2 + e 3 is the only way to decompose 1 into a sum of three orthogonal central idempotents. [This is useful since it is hard to tell when an idempotent is primitive.] Anyway, the decomposition 1 = e 1 + e 2 + e 3 gives a direct sum decomposition of any A 123 -module into summands: M = e 1 M e 2 M e 3 M where e 1 M is a module over the first block of A 123, e 2 M is a module over the second block and similarly for e 3 M. Note that e i is the projection of M onto e i M. The kernel is the other two summands since e i (e j M) = by orthogonality. 3 I 3

4 33.2 Example: G = S 3 Now let s look at the example G = S 3. This group has elements: 1, (12), (13), (23), (123), (132). }{{}}{{} conjugate conjugate These elements form three conjugacy classes as shown. Thus the group algebra has r = 3 blocks by Theorem By Theorem 33.2 these blocks must have size f i = 1, 1, 2 since = is the only way to write as a sum of three nonzero squares. By the discussion above this group algebra must have 3 primitive orthogonal central idempotents. They are: e 1 = 1 (1 + (123) + (132) + (12) + (13) + (23)) e 2 = 1 (1 + (123) + (132) (12) (13) (23)) e 3 = (123) 2 (132) The first idempotent occurs in every group algebra: e 1 = 1 This is always an idempotent. To see that e 2 is an idempotent write g g G a = 1 + (123) + (132) b = (12) + (13) + (23) Then a 2 = 3a [so 1 3 a is an idempotent], b2 = 3a and ab = 3b. Thus (a ± b) 2 = (a ± b) and e 2 = 1 (a b) is also idempotent. The last idempotent comes from the formula: e 3 = 1 e 1 e 2. [Since a = e 1 + e 2 is idempotent, so is 1 a: (1 a) 2 = 1 2a + a = 1 a.] Why are the numbers 1, 1, 4 in boldface? The reason is that they represent the dimensions of the simple subalgebras e 1 CG, e 2 CG and e 3 CG, respectively. This in turn follows from Theorem Theorem The coefficient of 1 in a central idempotent e is equal to 1 times the dimension of the subalgebra ecg. 4

5 Proof. Multiplication by e is the projection of CG to ecg. Therefore, ( ) dim ecg = Tr(e) = Tr ag g = a g Tr(g). Taking G as a basis for the vector space CG we see that if g 1 then Tr(g) =. And Tr(1) = dim CG =. Consequently, the trace of e as a linear endomorphism of CG [the regular representation] is a 1. Thus: dim ecg = Tr(e) = a 1 a 1 = dim ecg Characters at last At the end of the lecture, I finally decided I had better talk about characters. Definition If V is a G-module then the character of V is the function given by χ V (g) = Tr(g : V V ). χ V : G C The first theorem I proved about characters was the following theorem about the average value of the character of V. Theorem 33.. The dimension of the fixed submodule V G is equal to the average value of χ V (g), i.e., dim V G = 1 χ V (g) g G Proof. Recall that the projection map π : V V G is given by π(v) = 1 gv. In other words, π is multiplication by e 1 = 1 g on V. So: dim V G = Tr(π) = χ V (e 1 ) = 1 χv (g). This formula is an example of a more general formula about the dimension of Hom G (V, W ). In this case we are looking at V G = Hom G (C, V ) (1) where C is the trivial 1 dimensional representation given by C with the trivial action gz = z for all z C. Homomorphisms f : C V send 1 C to v V so that gv = gf(1) = f(g1) = f(1) = v, i.e., v V G. This proves (1). 5

6 Definition The inner product of characters χ V, χ W is defined by: χ V, χ W = g G χ V (g)χ W (g) where the overline denote complex conjugation. For example, the character of the trivial representation is χc(g) = 1 for all g G so χc, χ V = g G χ V (g) = dim V G. (2) Note that dimension is always real and thus equal to its own conjugate so we can ignore the stupid overline. This equation (1) is a special case of the following formula. Theorem Given G-modules V, W the G-module homomorphisms V W form a vector space of dimension dim Hom G (V, W ) = χ V, χ W At this point we ran out of time. However, I said that this follows from Theorem 33. and the formula: Hom G (V, W ) = Hom(V, W ) G.

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